With: 2mg=mv3^2/r-mv1^2/r
V3 being the speed at the top and V1 being the speed given at the bottom
You get: V3^2=2gr+V1^2
sqrt(2*9.8*3.1+16^2)
sqrt(316.76)=17.8
For the normal force in Figure 5.21 to have the same magnitude at all points on the vertical track, the stunt driver must adjust the speed to be different at different points. Suppose, for example, that the track has a radius of 3.1 m and that the driver goes past point 1 at the bottom with a speed of 16 m/s. What speed must she have at point 3, so that the normal force at the top has the same magnitude as it did at the bottom?
Is there some magic potion I can drink to reveal where Point 3 is?
Point 3 is located at the top.
force on bottom is mv^2/r+ mg. Calculate that.
Then at the top, it is mv^2/r - mg
set that equal to the bottom force, and solve for v
1 answer
1 answer